How to calculate rotation quaternion between two orientation quaternions?












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I have some device (3D pointer) connected to my computer which returns it's position (in cartesian XYZ system) and orientation (in quaternions). I receive this values about 30 times/sec.



Now I need to draw a figure on a screen that has the same position and orientation that my pointer. The position is simple, I can directly pass X,Y and Z to my software.



The problem is that the software doesn't accept orientation (a state) but only rotation (an operation). So I think I should calculate a difference between current and previous orientation - this would be my rotation operation.



Any ideas on how to do this? What quaternion operations should be done to get rotation out of two consecutive orientations?






3d rotations quaternions orientation







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    0












    $begingroup$


    I have some device (3D pointer) connected to my computer which returns it's position (in cartesian XYZ system) and orientation (in quaternions). I receive this values about 30 times/sec.



    Now I need to draw a figure on a screen that has the same position and orientation that my pointer. The position is simple, I can directly pass X,Y and Z to my software.



    The problem is that the software doesn't accept orientation (a state) but only rotation (an operation). So I think I should calculate a difference between current and previous orientation - this would be my rotation operation.



    Any ideas on how to do this? What quaternion operations should be done to get rotation out of two consecutive orientations?






    3d rotations quaternions orientation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have some device (3D pointer) connected to my computer which returns it's position (in cartesian XYZ system) and orientation (in quaternions). I receive this values about 30 times/sec.



      Now I need to draw a figure on a screen that has the same position and orientation that my pointer. The position is simple, I can directly pass X,Y and Z to my software.



      The problem is that the software doesn't accept orientation (a state) but only rotation (an operation). So I think I should calculate a difference between current and previous orientation - this would be my rotation operation.



      Any ideas on how to do this? What quaternion operations should be done to get rotation out of two consecutive orientations?






      3d rotations quaternions orientation







      share|cite|improve this question









      $endgroup$




      I have some device (3D pointer) connected to my computer which returns it's position (in cartesian XYZ system) and orientation (in quaternions). I receive this values about 30 times/sec.



      Now I need to draw a figure on a screen that has the same position and orientation that my pointer. The position is simple, I can directly pass X,Y and Z to my software.



      The problem is that the software doesn't accept orientation (a state) but only rotation (an operation). So I think I should calculate a difference between current and previous orientation - this would be my rotation operation.



      Any ideas on how to do this? What quaternion operations should be done to get rotation out of two consecutive orientations?






      3d rotations quaternions orientation




      3d rotations quaternions orientation






      share|cite|improve this question













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      asked May 12 '16 at 11:18









      crooveckcrooveck

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          $begingroup$

          If $$mathbf{q_1} = w_1 + x_1mathbf{i} + y_1mathbf{j} + z_1mathbf{k}$$ is an unit quaternion, $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$representing the first (earlier) rotation, and $$mathbf{q_2} = w_2 + x_2mathbf{i} + y_2mathbf{j} + z_2mathbf{k}$$ is a second (later) rotation, then their Hamilton product represents the combined rotation,
          $$begin{align}mathbf{q} = mathbf{q_2} mathbf{q_1} = ; & w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\
          + ; & (w_1 x_2 + w_2 x_1 - y_1 z_2 + y_2 z_1) ; mathbf{i}\
          + ; & (w_1 y_2 + w_2 y_1 + x_1 z_2 - x_2 z_1) ; mathbf{j}\
          + ; & (w_1 z_2 + w_2 z_1 - x_1 y_2 + x_2 y_1) ; mathbf{k}
          end{align}$$
          Note that the Hamilton product is not commutative; the order of the quaternions matters. The result is also an unit quaternion, except for any numerical errors that might creep in. Fortunately, you can always normalize the rotation quaternion,
          $$mathbf{q'} = frac{w + x ;mathbf{i} + y ;mathbf{j} + z ;mathbf{k}}{sqrt{w^2 + x^2 + y^2 + z^2}}$$
          to avoid compounding errors. (It is basically safe to do after each operation, but usually necessary only after a few products.)



          The inverse of a rotation is
          $$mathbf{q}^{-1} = w - x ;mathbf{i} - y ;mathbf{j} - z ;mathbf{k}$$
          ie. negating all components of an unit quaternion, except for the scalar component $w$, inverts the rotation. (Negating all components does not change the rotation it represents.)



          You can also interpolate between two unit quaternions, $0 le p le 1$,
          $$begin{align}mathbf{q'} = (1-p)mathbf{q_1} + pmathbf{q_2} & = w_1 + p (w_2 - w_1)\
          & + left( x_1 + p (x_2 - x_1) right ) ; mathbf{i} \
          & + left( y_1 + p (y_2 - y_1) right ) ; mathbf{j} \
          & + left( z_1 + p (z_2 - z_1) right ) ; mathbf{k}end{align}$$
          if you normalize the result to unit length,
          $$mathbf{q} = frac{w + x' ;mathbf{i} + y' ;mathbf{j} + z' ;mathbf{k}}{sqrt{w'^2 + x'^2 + y'^2 + z'^2}}$$
          This is very useful in camera movement between two orientations. To get a really smooth change, use e.g.
          $$p = 3 p'^2 - 2 p'^3$$
          or
          $$p = 6 p'^5 - 15 p'^4 + 10 p'^3$$
          with $0 le p' le 1$. Both start and stop with zero velocity, but the former has a fixed rate of change of acceleration ("jerk"), and the latter starts and stops with zero acceleration.






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            $begingroup$

            If $$mathbf{q_1} = w_1 + x_1mathbf{i} + y_1mathbf{j} + z_1mathbf{k}$$ is an unit quaternion, $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$representing the first (earlier) rotation, and $$mathbf{q_2} = w_2 + x_2mathbf{i} + y_2mathbf{j} + z_2mathbf{k}$$ is a second (later) rotation, then their Hamilton product represents the combined rotation,
            $$begin{align}mathbf{q} = mathbf{q_2} mathbf{q_1} = ; & w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\
            + ; & (w_1 x_2 + w_2 x_1 - y_1 z_2 + y_2 z_1) ; mathbf{i}\
            + ; & (w_1 y_2 + w_2 y_1 + x_1 z_2 - x_2 z_1) ; mathbf{j}\
            + ; & (w_1 z_2 + w_2 z_1 - x_1 y_2 + x_2 y_1) ; mathbf{k}
            end{align}$$
            Note that the Hamilton product is not commutative; the order of the quaternions matters. The result is also an unit quaternion, except for any numerical errors that might creep in. Fortunately, you can always normalize the rotation quaternion,
            $$mathbf{q'} = frac{w + x ;mathbf{i} + y ;mathbf{j} + z ;mathbf{k}}{sqrt{w^2 + x^2 + y^2 + z^2}}$$
            to avoid compounding errors. (It is basically safe to do after each operation, but usually necessary only after a few products.)



            The inverse of a rotation is
            $$mathbf{q}^{-1} = w - x ;mathbf{i} - y ;mathbf{j} - z ;mathbf{k}$$
            ie. negating all components of an unit quaternion, except for the scalar component $w$, inverts the rotation. (Negating all components does not change the rotation it represents.)



            You can also interpolate between two unit quaternions, $0 le p le 1$,
            $$begin{align}mathbf{q'} = (1-p)mathbf{q_1} + pmathbf{q_2} & = w_1 + p (w_2 - w_1)\
            & + left( x_1 + p (x_2 - x_1) right ) ; mathbf{i} \
            & + left( y_1 + p (y_2 - y_1) right ) ; mathbf{j} \
            & + left( z_1 + p (z_2 - z_1) right ) ; mathbf{k}end{align}$$
            if you normalize the result to unit length,
            $$mathbf{q} = frac{w + x' ;mathbf{i} + y' ;mathbf{j} + z' ;mathbf{k}}{sqrt{w'^2 + x'^2 + y'^2 + z'^2}}$$
            This is very useful in camera movement between two orientations. To get a really smooth change, use e.g.
            $$p = 3 p'^2 - 2 p'^3$$
            or
            $$p = 6 p'^5 - 15 p'^4 + 10 p'^3$$
            with $0 le p' le 1$. Both start and stop with zero velocity, but the former has a fixed rate of change of acceleration ("jerk"), and the latter starts and stops with zero acceleration.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              If $$mathbf{q_1} = w_1 + x_1mathbf{i} + y_1mathbf{j} + z_1mathbf{k}$$ is an unit quaternion, $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$representing the first (earlier) rotation, and $$mathbf{q_2} = w_2 + x_2mathbf{i} + y_2mathbf{j} + z_2mathbf{k}$$ is a second (later) rotation, then their Hamilton product represents the combined rotation,
              $$begin{align}mathbf{q} = mathbf{q_2} mathbf{q_1} = ; & w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\
              + ; & (w_1 x_2 + w_2 x_1 - y_1 z_2 + y_2 z_1) ; mathbf{i}\
              + ; & (w_1 y_2 + w_2 y_1 + x_1 z_2 - x_2 z_1) ; mathbf{j}\
              + ; & (w_1 z_2 + w_2 z_1 - x_1 y_2 + x_2 y_1) ; mathbf{k}
              end{align}$$
              Note that the Hamilton product is not commutative; the order of the quaternions matters. The result is also an unit quaternion, except for any numerical errors that might creep in. Fortunately, you can always normalize the rotation quaternion,
              $$mathbf{q'} = frac{w + x ;mathbf{i} + y ;mathbf{j} + z ;mathbf{k}}{sqrt{w^2 + x^2 + y^2 + z^2}}$$
              to avoid compounding errors. (It is basically safe to do after each operation, but usually necessary only after a few products.)



              The inverse of a rotation is
              $$mathbf{q}^{-1} = w - x ;mathbf{i} - y ;mathbf{j} - z ;mathbf{k}$$
              ie. negating all components of an unit quaternion, except for the scalar component $w$, inverts the rotation. (Negating all components does not change the rotation it represents.)



              You can also interpolate between two unit quaternions, $0 le p le 1$,
              $$begin{align}mathbf{q'} = (1-p)mathbf{q_1} + pmathbf{q_2} & = w_1 + p (w_2 - w_1)\
              & + left( x_1 + p (x_2 - x_1) right ) ; mathbf{i} \
              & + left( y_1 + p (y_2 - y_1) right ) ; mathbf{j} \
              & + left( z_1 + p (z_2 - z_1) right ) ; mathbf{k}end{align}$$
              if you normalize the result to unit length,
              $$mathbf{q} = frac{w + x' ;mathbf{i} + y' ;mathbf{j} + z' ;mathbf{k}}{sqrt{w'^2 + x'^2 + y'^2 + z'^2}}$$
              This is very useful in camera movement between two orientations. To get a really smooth change, use e.g.
              $$p = 3 p'^2 - 2 p'^3$$
              or
              $$p = 6 p'^5 - 15 p'^4 + 10 p'^3$$
              with $0 le p' le 1$. Both start and stop with zero velocity, but the former has a fixed rate of change of acceleration ("jerk"), and the latter starts and stops with zero acceleration.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                If $$mathbf{q_1} = w_1 + x_1mathbf{i} + y_1mathbf{j} + z_1mathbf{k}$$ is an unit quaternion, $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$representing the first (earlier) rotation, and $$mathbf{q_2} = w_2 + x_2mathbf{i} + y_2mathbf{j} + z_2mathbf{k}$$ is a second (later) rotation, then their Hamilton product represents the combined rotation,
                $$begin{align}mathbf{q} = mathbf{q_2} mathbf{q_1} = ; & w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\
                + ; & (w_1 x_2 + w_2 x_1 - y_1 z_2 + y_2 z_1) ; mathbf{i}\
                + ; & (w_1 y_2 + w_2 y_1 + x_1 z_2 - x_2 z_1) ; mathbf{j}\
                + ; & (w_1 z_2 + w_2 z_1 - x_1 y_2 + x_2 y_1) ; mathbf{k}
                end{align}$$
                Note that the Hamilton product is not commutative; the order of the quaternions matters. The result is also an unit quaternion, except for any numerical errors that might creep in. Fortunately, you can always normalize the rotation quaternion,
                $$mathbf{q'} = frac{w + x ;mathbf{i} + y ;mathbf{j} + z ;mathbf{k}}{sqrt{w^2 + x^2 + y^2 + z^2}}$$
                to avoid compounding errors. (It is basically safe to do after each operation, but usually necessary only after a few products.)



                The inverse of a rotation is
                $$mathbf{q}^{-1} = w - x ;mathbf{i} - y ;mathbf{j} - z ;mathbf{k}$$
                ie. negating all components of an unit quaternion, except for the scalar component $w$, inverts the rotation. (Negating all components does not change the rotation it represents.)



                You can also interpolate between two unit quaternions, $0 le p le 1$,
                $$begin{align}mathbf{q'} = (1-p)mathbf{q_1} + pmathbf{q_2} & = w_1 + p (w_2 - w_1)\
                & + left( x_1 + p (x_2 - x_1) right ) ; mathbf{i} \
                & + left( y_1 + p (y_2 - y_1) right ) ; mathbf{j} \
                & + left( z_1 + p (z_2 - z_1) right ) ; mathbf{k}end{align}$$
                if you normalize the result to unit length,
                $$mathbf{q} = frac{w + x' ;mathbf{i} + y' ;mathbf{j} + z' ;mathbf{k}}{sqrt{w'^2 + x'^2 + y'^2 + z'^2}}$$
                This is very useful in camera movement between two orientations. To get a really smooth change, use e.g.
                $$p = 3 p'^2 - 2 p'^3$$
                or
                $$p = 6 p'^5 - 15 p'^4 + 10 p'^3$$
                with $0 le p' le 1$. Both start and stop with zero velocity, but the former has a fixed rate of change of acceleration ("jerk"), and the latter starts and stops with zero acceleration.






                share|cite|improve this answer









                $endgroup$



                If $$mathbf{q_1} = w_1 + x_1mathbf{i} + y_1mathbf{j} + z_1mathbf{k}$$ is an unit quaternion, $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$representing the first (earlier) rotation, and $$mathbf{q_2} = w_2 + x_2mathbf{i} + y_2mathbf{j} + z_2mathbf{k}$$ is a second (later) rotation, then their Hamilton product represents the combined rotation,
                $$begin{align}mathbf{q} = mathbf{q_2} mathbf{q_1} = ; & w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\
                + ; & (w_1 x_2 + w_2 x_1 - y_1 z_2 + y_2 z_1) ; mathbf{i}\
                + ; & (w_1 y_2 + w_2 y_1 + x_1 z_2 - x_2 z_1) ; mathbf{j}\
                + ; & (w_1 z_2 + w_2 z_1 - x_1 y_2 + x_2 y_1) ; mathbf{k}
                end{align}$$
                Note that the Hamilton product is not commutative; the order of the quaternions matters. The result is also an unit quaternion, except for any numerical errors that might creep in. Fortunately, you can always normalize the rotation quaternion,
                $$mathbf{q'} = frac{w + x ;mathbf{i} + y ;mathbf{j} + z ;mathbf{k}}{sqrt{w^2 + x^2 + y^2 + z^2}}$$
                to avoid compounding errors. (It is basically safe to do after each operation, but usually necessary only after a few products.)



                The inverse of a rotation is
                $$mathbf{q}^{-1} = w - x ;mathbf{i} - y ;mathbf{j} - z ;mathbf{k}$$
                ie. negating all components of an unit quaternion, except for the scalar component $w$, inverts the rotation. (Negating all components does not change the rotation it represents.)



                You can also interpolate between two unit quaternions, $0 le p le 1$,
                $$begin{align}mathbf{q'} = (1-p)mathbf{q_1} + pmathbf{q_2} & = w_1 + p (w_2 - w_1)\
                & + left( x_1 + p (x_2 - x_1) right ) ; mathbf{i} \
                & + left( y_1 + p (y_2 - y_1) right ) ; mathbf{j} \
                & + left( z_1 + p (z_2 - z_1) right ) ; mathbf{k}end{align}$$
                if you normalize the result to unit length,
                $$mathbf{q} = frac{w + x' ;mathbf{i} + y' ;mathbf{j} + z' ;mathbf{k}}{sqrt{w'^2 + x'^2 + y'^2 + z'^2}}$$
                This is very useful in camera movement between two orientations. To get a really smooth change, use e.g.
                $$p = 3 p'^2 - 2 p'^3$$
                or
                $$p = 6 p'^5 - 15 p'^4 + 10 p'^3$$
                with $0 le p' le 1$. Both start and stop with zero velocity, but the former has a fixed rate of change of acceleration ("jerk"), and the latter starts and stops with zero acceleration.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 13 '16 at 23:20









                Nominal AnimalNominal Animal

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